Mixing
Prove if wronskian is zero then solutions of ODE are linearly dependant
$begingroup$
Suppose we have $$y'' + P(x)y' + Q(x)y = 0 $$
and let the solutions be $y_1, y_2$ in $[a,b]$. The wronskian is $y_1 y_2' - y_2 y_1'$, which is identically zero. Now lets consider 3 cases:
If $y_1$ is identically zero, then so is $W(x)$ and so $y_1 = 0(y_2)$ so they are linearly dependant.
If $y_1$ is never zero in $[a,b]$ then $frac{W(x)}{y_1^2} = left(frac{y_2}{y_1}right)' = 0$ or $y_2 = c y_1$ so they are L.D.
What if $y_1$ is zero only on some points in $[a,b]$ and say $y_1'$ is also zero at those points, how do we show $y_2, y_2'$ is also zero at those points
calculus ordinary-differential-equations
asked Jan 11 at 2:13
jeeajeea
57515
$endgroup$
add a comment |
$begingroup$
Suppose we have $$y'' + P(x)y' + Q(x)y = 0 $$
and let the solutions be $y_1, y_2$ in $[a,b]$. The wronskian is $y_1 y_2' - y_2 y_1'$, which is identically zero. Now lets consider 3 cases:
If $y_1$ is identically zero, then so is $W(x)$ and so $y_1 = 0(y_2)$ so they are linearly dependant.
If $y_1$ is never zero in $[a,b]$ then $frac{W(x)}{y_1^2} = left(frac{y_2}{y_1}right)' = 0$ or $y_2 = c y_1$ so they are L.D.
What if $y_1$ is zero only on some points in $[a,b]$ and say $y_1'$ is also zero at those points, how do we show $y_2, y_2'$ is also zero at those points
calculus ordinary-differential-equations
asked Jan 11 at 2:13
jeeajeea
57515
$endgroup$
add a comment |
$begingroup$
Suppose we have $$y'' + P(x)y' + Q(x)y = 0 $$
and let the solutions be $y_1, y_2$ in $[a,b]$. The wronskian is $y_1 y_2' - y_2 y_1'$, which is identically zero. Now lets consider 3 cases:
If $y_1$ is identically zero, then so is $W(x)$ and so $y_1 = 0(y_2)$ so they are linearly dependant.
If $y_1$ is never zero in $[a,b]$ then $frac{W(x)}{y_1^2} = left(frac{y_2}{y_1}right)' = 0$ or $y_2 = c y_1$ so they are L.D.
What if $y_1$ is zero only on some points in $[a,b]$ and say $y_1'$ is also zero at those points, how do we show $y_2, y_2'$ is also zero at those points
calculus ordinary-differential-equations
asked Jan 11 at 2:13
jeeajeea
57515
$endgroup$
Suppose we have $$y'' + P(x)y' + Q(x)y = 0 $$
and let the solutions be $y_1, y_2$ in $[a,b]$. The wronskian is $y_1 y_2' - y_2 y_1'$, which is identically zero. Now lets consider 3 cases:
If $y_1$ is identically zero, then so is $W(x)$ and so $y_1 = 0(y_2)$ so they are linearly dependant.
If $y_1$ is never zero in $[a,b]$ then $frac{W(x)}{y_1^2} = left(frac{y_2}{y_1}right)' = 0$ or $y_2 = c y_1$ so they are L.D.
What if $y_1$ is zero only on some points in $[a,b]$ and say $y_1'$ is also zero at those points, how do we show $y_2, y_2'$ is also zero at those points
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 11 at 2:13
jeeajeea
57515
asked Jan 11 at 2:13
jeeajeea
57515
asked Jan 11 at 2:13
jeeajeea
57515
asked Jan 11 at 2:13
jeeajeea
57515
asked Jan 11 at 2:13
jeeajeea
57515
57515
add a comment |
add a comment |
1 Answer
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$begingroup$
If the list ${y_1, y_2 } $ were independent, then we can find nontrivial $(c_1, c_2)$ so that
$$ c_1 y_1 + c_2 y_2 = 0$$
differentiation gives
$$ c_1 y_1' + c_2 y_2' = 0 $$
and so the system
$$ begin{bmatrix} y_1 & y_2 \ y_1' & y_2' end{bmatrix} begin{bmatrix} c_1 \ c_2 end{bmatrix} = 0 $$
has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If the list ${y_1, y_2 } $ were independent, then we can find nontrivial $(c_1, c_2)$ so that
$$ c_1 y_1 + c_2 y_2 = 0$$
differentiation gives
$$ c_1 y_1' + c_2 y_2' = 0 $$
and so the system
$$ begin{bmatrix} y_1 & y_2 \ y_1' & y_2' end{bmatrix} begin{bmatrix} c_1 \ c_2 end{bmatrix} = 0 $$
has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
$endgroup$
add a comment |
$begingroup$
If the list ${y_1, y_2 } $ were independent, then we can find nontrivial $(c_1, c_2)$ so that
$$ c_1 y_1 + c_2 y_2 = 0$$
differentiation gives
$$ c_1 y_1' + c_2 y_2' = 0 $$
and so the system
$$ begin{bmatrix} y_1 & y_2 \ y_1' & y_2' end{bmatrix} begin{bmatrix} c_1 \ c_2 end{bmatrix} = 0 $$
has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
$endgroup$
add a comment |
$begingroup$
If the list ${y_1, y_2 } $ were independent, then we can find nontrivial $(c_1, c_2)$ so that
$$ c_1 y_1 + c_2 y_2 = 0$$
differentiation gives
$$ c_1 y_1' + c_2 y_2' = 0 $$
and so the system
$$ begin{bmatrix} y_1 & y_2 \ y_1' & y_2' end{bmatrix} begin{bmatrix} c_1 \ c_2 end{bmatrix} = 0 $$
has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
$endgroup$
If the list ${y_1, y_2 } $ were independent, then we can find nontrivial $(c_1, c_2)$ so that
$$ c_1 y_1 + c_2 y_2 = 0$$
differentiation gives
$$ c_1 y_1' + c_2 y_2' = 0 $$
and so the system
$$ begin{bmatrix} y_1 & y_2 \ y_1' & y_2' end{bmatrix} begin{bmatrix} c_1 \ c_2 end{bmatrix} = 0 $$
has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
answered Jan 11 at 2:21
Jimmy SabaterJimmy Sabater
2,629321
2,629321
add a comment |
add a comment |
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